Question

Acetyl CoA turns the citric acid cycle one turn. How many turns will the citric acid cycle turn when 57.3 g of glucose (C6H1206) is oxidized in the cells?
O 125 x 1024 turns
O
3.83 x 1023 turns
O
2.94 x 1023
turns
O
9.36 x 1024 turns
O 5.71 x 1023 tums
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Answer 1

The number of turns that citric acid cycle will turn when 57.3 g of glucose (C6H1206) is oxidized in the cells is 1.92 × 10²³ turns.

How to calculate number of turns?

According to this question, Acetyl CoA turns the citric acid cycle one turn.

First, we find the number of moles of 57.3 grams of glucose by dividing by its molar mass of 180g/mol.

no of moles = 57.3/180 = 0.32moles

Next, we multiply the number of moles by Avogadro's number i.e. 6.02 × 10²³ molecules.

no of turns = 6.02 × 10²³ × 0.32

no of turns = 1.92 × 10²³ turns

Therefore, the number of turns that citric acid cycle will turn when 57.3 g of glucose (C6H1206) is oxidized in the cells is 1.92 × 10²³ turns.

Learn more about no of turns at: brainly.com/question/1747943

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